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Andreyy89
3 years ago
10

Barium nitrate has the formula Ba( NO3)2. Which statement is true about barium nitrate?

Chemistry
1 answer:
OLga [1]3 years ago
4 0
I found these four statements for that question:

Each molecule contains four different elements.

Each molecule contains three atoms.

Each molecule contains seven different bonds.

Each molecule contains six oxygen atoms.

The last one is true. Each molecule contains six oxygen atoms.
The number to the right of O and of (NO3) ares subscripts.


The chemical formula uses subscripts to indicate the number of atoms.



The subscript 2 in (NO3)2 means that there are two NO3 radicals.


And the subscript 3 to the right of O means that each NO3 radical has three atoms of O.

Then, the number of atoms of O is 2 * 3 = 6.


So, the true statement is the last one: each molecule of Ba (NO3)2 has six atoms of O.


From that molecule you can also tell:

- Each molecule contains one atom of barium

- Each molecule contains two atoms of nitrogen

- Each molecule contains two NO3 radicals

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Colt1911 [192]

Answer:

<em>The Cloud of gas .</em>

Explanation:

<em>Once the pressure gets high enough, it begins to rival the strength of gravity and the collapse of the cloud slows down. Eventually, the cloud of gas becomes a protostar: an infant star that has not yet begun to fuse hydrogen in its core.</em>

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3 years ago
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julsineya [31]

Answer:

C.  Cs, because it has the lower ionization energy and more easily gives up its valence electrons to participate in a reaction.

Explanation:

The Cs will produce the most hydrogen gas because it has a lower ionization energy and more easily gives up its valence electrons to participate in the reaction.

  • Ionization energy is the energy required to remove the most loosely held electrons in an atom.
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3 years ago
ASAP NEED HELP PLSSS
lord [1]

Answer:

Volume = 5.73L

Explanation:

Data;

V1 = 3.75L

P1 = 0.980atm

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V2 = ?

This question involves the use of Boyle's law, which states that, the volume of a fixed mass of gas is inversely proportional to its pressure provided that temperature remains constant.

Mathematically,

V = kP, k = PV

P1V1 = P2V2 =P3V3=........=PnVn

P1V1 = P2V2

V2 = (P1 * V1) / P2

V2 = (0.980 × 3.75) / 0.641

V2 = 5.73L

The final volume of the gas is 5.73L

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What subatomic particle determines how an element reacts with another element?
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Hello my friend,
No. of valence electron determine the reaction between elements.
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3 years ago
A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo
atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

             Weight = 0.09 \times 16.04 + 0.91 \times 29

                          = 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

                    \frac{800}{27.8336}

                     = 28.74 kmol/hr

Now, applying component balance as follows.

                0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}

                   F_{p} = 43.11 kmol/hr

                   F_{A} + F_{B} = F_{p}

                          F_{B} = 43.11 - 28.74

                                      = 14.37 kmol/hr

So, mass flow rate of pure (B), is F_{B} = 14.37 \times 29

                                                    = 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

             Molecular weight of product stream = Mol. weight \times 43.11 kmol/hr

                                  = 0.06 \times 16.04 + 0.94 \times 29

                                  = 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of O_{2} into the product stream is as follows.

                    0.21 \times 0.94 \times 43.11

                      = 8.5099 kmol/hr \times 329 g/mol

                      = 272.31 kg/hr

Therefore, calculate the mass % of O_{2} into the stream as follows.

                 \frac{272.31}{1216.67} \times 100

                     = 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of O_{2} in the product gas is 22.38%.

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